Question

If $z=x+iy$, where $x$ and $y$ are real numbers and $i=\sqrt{-1}$, then the points $(x,y)$ for which $\frac{z-1}{z+1}$ is real, lie on

• A. an ellipse
• B. a circle
• C. a parabola
• D. a straight line

Answer 1

The correct option is D a straight line

Given $z=x+iy$
Now, $\frac{z-1}{z-i}=\frac{(x+iy)-1}{(x+iy)-i}$
$=\frac{(x-1)+iy}{x+i(y-1)}\times \frac{x-i(y-1)}{x-i(y-1)}$
$=\frac{x(x-1)+ixy-i(x-1)(y-1)+y(y-1)}{x^2+{(y-1)}^2}$
$\frac{(x^2+y^2-x-y)+i(xy-xy+y+x-1)}{(x^2+y^2+1^{-2y})}$
$=\left(\frac{x^2+y^2-x-y}{x^2+y^2-2y+1}\right)+i\left(\frac{x+y-1}{x^2+y^2-2y+1}\right)......\left(i\right)$
Givne $\frac{z-1}{z-i}$ is real
So, its imaginary part should be zero
ie., $\frac{x+y-1}{x^2+y^2-2y+1}=0$
$\Rightarrow$ $x+y=1$, ($x^2+y^2-2y+1\ne 0$)
which represent a straight line